3 and 4 .Determinants and Matrices
hard

સુરેખ સમીકરણ સંહિતા 

$(\lambda-1) x+(3 \lambda+1) y+2 \lambda z=0$

$(\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0$

$2 x+(3 \lambda+1) y+3(\lambda-1) z=0$

ને શુન્યેતર ઉકેલો હોય તો $\lambda$ ની બધી ભિન્ન કિમતોનો સરવાળો શોધો 

A

$3$

B

$0$

C

$6$

D

$9$

(JEE MAIN-2020)

Solution

$(\lambda-1) x+(3 \lambda+1) y+2 \lambda z=0$

$(\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0$

$2 x+(3 \lambda+1) y+(3 \lambda-3) z=0$

$\left|\begin{array}{ccc}\lambda-1 & 3 \lambda+1 & 2 \lambda \\ \lambda-1 & 4 \lambda-2 & \lambda+3 \\ 2 & 3 \lambda+1 & 3 \lambda-3\end{array}\right|=0$

$R _{1} \rightarrow R _{1}- R _{2}$ and $R _{2} \rightarrow R _{2}- R _{3}$

$\left|\begin{array}{ccc}0 & 3-\lambda & \lambda-3 \\ \lambda-3 & \lambda-3 & -2(\lambda-3) \\ 2 & 3 \lambda+1 & 3 \lambda-3\end{array}\right|=0$

$(\lambda-3)^{2}\left|\begin{array}{ccc}0 & -1 & 1 \\ 1 & 1 & -2 \\ 2 & 3 \lambda+1 & 3 \lambda-3\end{array}\right|=0$

$(\lambda-3)^{2}[(3 \lambda+1)+(3 \lambda-1)]=0$

$6 \lambda(\lambda-3)^{2}=0 \Rightarrow \lambda=0,3$

$\operatorname{Sum}=3$

Standard 12
Mathematics

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